Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
Question
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Chapter 13, Problem 10P
To determine

The minimum EUAC for the machine and its economic life.

Expert Solution & Answer
Check Mark

Answer to Problem 10P

The minimum EUAC is $17240.

The economic life is 4years.

Explanation of Solution

Given:

The first cost is $50000.

Decline in its value per annum is 20%.

Operating and maintenance cost is $3500 per year.

Climb in the operating and maintenance cost is $2000 per year.

The MARR is 9%.

Calculation:

Write the expression for capital recovery.

(A/P,i,n)=(i( 1+i)n( 1+i)n1) ...... (I)

Here, the present value is P, the annual expenditure is A, the rate of interest is i and the period is n.

Write the expression for uniform gradient series.

(A/G)=(1in( 1+i)n1) ...... (II)

Here, the gradient is G.

Calculate the Equivalent Uniform Annualised Cost of capital.

Year Fixed Cost (P) (a) Salvage percent (b)=0.80 Salvage Value (c)=(cn1×b) (A/P,9%,n) (d) EUAC ((ac)×d)+(c×0.09)
0 $50000 $50000
1 $50000 0.80 $40000 1.090 $14500
2 $50000 0.80 $32000 0.5685 $13113
3 $50000 0.80 $25600 0.3951 $11944.44
4 $50000 0.80 $20480 0.3087 $10956.024
5 $50000 0.80 $16384 0.2571 $10117.024
6 $50000 0.80 $13107.2 0.2229 $9403.05312
7 $50000 0.80 $10485.76 0.1987 $8795.19788
8 $50000 0.80 $8388.608 0.1807 $8274.153254
9 $50000 0.80 $6710.8864 0.1668 $7824.603924
10 $50000 0.80 $5368.70912 0.1558 $7437.73894

Now the annual cost is and the climb in cost is given.

This means that the provided cash flow will be of the uniform gradient series where $2000 will act as G and $3500 as A.

Calculate the equal uniform annualised cost of machine.

Year A (a) G (b) (A/G,9%,n) (c) EUAC (a+(b×c))
0
1 $3500 $2000 0 $3500
2 $3500 $2000 0.478 $4456
3 $3500 $2000 0.943 $5386
4 $3500 $2000 1.393 $6286
5 $3500 $2000 1.828 $7156
6 $3500 $2000 2.25 $8000
7 $3500 $2000 2.657 $8814
8 $3500 $2000 3.051 $9602
9 $3500 $2000 3.431 $10362
10 $3500 $2000 3.798 $11096

Calculate the total equivalent uniform annualized cost.

Year EUAC of capital (a) EUAC of machine (b) Total EUAC c=(a+b)
1 $14500 $3500 $18000
2 $13113 $4456 $17569
3 $11944.44 $5386 $17330.44
4 $10956.024 $6286 $17242.024
5 $10117.024 $7156 $17273.233
6 $9403.05312 $8000 $17403.053120
7 $8795.19788 $8814 $17609.197888
8 $8274.153254 $9602 $17876.153254
9 $7824.603924 $10362 $18186.603924
10 $7437.73894 $11096 $18532.738940

From the table it is clearly known that the minimum EUAC is $17242.024$17240 and the economic life is 4years.

Conclusion:

The minimum EUAC is $17240.

The economic life is 4years.

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